Three particles,each having a charge of $10 \mu C$,are placed at the corners of an equilateral triangle of side $10 \ cm$. The electrostatic potential energy of the system is (Given $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \ N \cdot m^{2}/C^{2}$):

An electric dipole is formed by two charges $+q$ and $-q$ located in the $xy$-plane at $(0, 2) \text{ mm}$ and $(0, -2) \text{ mm}$,respectively,as shown in the figure. The electric potential at point $P(100, 100) \text{ mm}$ due to the dipole is $V_0$. The charges $+q$ and $-q$ are then moved to the points $(-1, 2) \text{ mm}$ and $(1, -2) \text{ mm}$,respectively. What is the value of the electric potential at $P$ due to the new dipole?

Three charges,each of magnitude $3 \mu C$,are placed on the vertices of an equilateral triangle of side $6 \ cm$. The net potential energy of the system will be nearly $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \ SI \ unit\right]$. (in $J$)

Three charges $Q$,$+q$,and $+q$ are placed at the vertices of a right-angled triangle as shown. The net electrostatic potential energy of the configuration is zero. The value of $Q$ is

Three point charges $+q$, $+2q$, and $+Q$ are placed at the three vertices of an equilateral triangle. If the potential energy of the system of three charges is zero, the value of $Q$ in terms of $q$ is

In the figure shown,the electric potential energy of the system is: ($q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$)